 Hi, I'm Zor. Welcome to Unisor Education. I would like to solve a couple of problems related to Lorentz force. This lecture is part of the course called Physics for Teens, presented on Unisor.com. There is, by the way, a prerequisite course called Math for Teens. Physics is using lots of math, so you really have to know math. So if you need Math for Teens is for you as well. Now I would recommend you to watch this lecture from this website rather than from, let's say, YouTube or wherever you found it. Because the site contains very detailed notes for each lecture. It's like a textbook, basically. Plus, if you would like to register, you can save some information about your exams. Maybe your parents or teachers can actually get involved in your education, basically assigning you certain things, certain topics from the course, etc. But it's not necessary. You can go to the website. It's completely free. There are no ads. You don't have to sign in and there are even exams if you want to take it. So we have three problems related to the Lorentz force. So the problem number one is the following. Consider there is a uniform magnetic field. So this is north and south of some kind of a magnet which is around it. But in between these two, it's a uniform. For instance, it can be, you know, the shoe, the horse shoe kind of a magnet. This is a north, this is a south. And this is the lines of the field which are, we can assume, it's uniform or any other way of creating the uniform magnetic field. Now, there is some kind of a wire here hanging on two threads, let's say. Now this is, I tried to to picture it in three dimensions, but basically for my purpose it's much more important to show it in such a way that the wire is like this pen, basically. This way. To show how it behaves in the magnetic field. So this is my wire. It goes through this board perpendicular to the board. These are magnetic fields, field lines. Okay, so what happens if there is an electric current going through this particular wire? Well, we have the Lorentz force, right? So there is a Lorentz force and it forces the the whole wire to basically move perpendicularly to both direction of the magnetic force and direction of the current. Right? So let's assume that it goes this way. Now, but since it's hanging, so these are the hanging needs, it will basically take this position. So from this position, it will go to this position. It has certain mass. The wire has certain mass. So there is a gravity. We are assuming this is in the gravitational field of Earth. So there is some kind of an acceleration of the free fall. So it cannot really go all the way up. Because, even if the magnet is large enough, let's consider it's still within the field, but it will still not go all the way up. Well, it depends, obviously, how strong the field is. But it will go a little bit up and then the force of the gravity will push it back here, right? So we have two different forces. The Lorentz force, which goes always in this particular direction. And we have a here, we have a gravity here. This is gravity, and this is the Lorentz force. And obviously there is a tension of the wire. Now at a certain moment it will stop because these three forces will balance each other. Now here is the problem. So there is an s, there is a g, there is an angle phi, at which it actually stopped and it reaches the equilibrium. These three forces are in equilibrium. Now what else? Now this particular wire has certain lengths and there is a current which goes through it. Now in my problem, I don't really have the current, but I have the voltage and the resistance of the wire. So this this wire is connected to a certain source of electricity which has the voltage and the wire has certain resistance. So these are all given. Now what remains to be determined is what is the force of the magnetic field. Remember Tesla? These are units. We are measuring the intensity of the magnetic field, right? So that's what we have to find out. We have to measure intensity of the field. Now obviously we will use the Lorentz force, which basically states that this Lorentz force is equal to the current times lengths of the wire times the intensity of the magnetic force. So we were measuring the force, the current and the lengths and we have determined that the force is proportional to these two things. And this is basically a coefficient of proportionality, which depends on the field. Obviously the stronger the field, magnetic field, the greater will be the force with given current and the lengths. So this equation can be a measurement. If we know the force, it can be a measurement of the magnetic field's intensity. And that's exactly what we would like to find out here. We don't have the force. However, we know that these three forces T, F and this is Mg. I think I use capital M in my problem. P is equal to Mg. So we know that these three forces are in equilibrium to each other. So that gives us the force. We can determine the force. And from the force, we will determine the magnetic field's intensity. Okay? So let's do first the mechanical part to find the force. And then we will do the electromagnetic part from the force. We will find the intensity of the field. So let me just take this down. This used to be M. So this is our wire and this is the angle it actually moves from the vertical direction under the influence of the Lorentz force. So this is the Lorentz force F. Now this is the weight Mg. And this is the tension. Now, and we know the angle phi, all right? And we know the mass, obviously. Well, that's what the force, the weight actually is. Okay, so how can we basically make all the these in equilibrium? This is how. So this is T. Then to the left, this is phi. Goes T times T times sine of phi. This is T times sine of phi. And this is T times cosine of phi. Now, if this thing is in balance, in equilibrium, the T sine phi is nullifying the F and T cosine is nullifying the weight P, right? So these two are equal and these two are equal in absolute value. And obviously the direction is opposite. And that's why we have complete equilibrium. Well, that's basically enough to find the force. Here it is. So first, T is equal to Mg divided by cosine phi. And if we will multiply it by sine, it will be the force phi. The force, the Lorentz force F. We're talking about absolute value, right? So don't let's not use the vectors here. I mean, we know about the vectors. Vectors are opposite. So we're talking about absolute value, all right? So this is the value of the F, the Lorentz force. Now, from this, we can find the intensity of the magnetic field, right? So B is equal to F divided by I l. Now we can obviously use the law that I, the current, is equal to voltage divided by resistance, the Ohm's law, right? So it's the voltage. Resistance goes here way and this is l. So from this we can find the B is equal to, instead of F, we put this. Now, by the way, this is tangent phi. Hope you remember the trigonometry. So it's Mg tangent phi. This is F. So times r and divided by u, l. So all these variables are used in this formula and this is the intensity of the field. Now let's just think about it and how logical and reasonable the formula actually is. Obviously, if, given the angle, the more massive our wire is, the more force we need to, more Lorentz force, which means the more intensity of the field we need to move this mass to this angle, right? Similarly, if angle is given, if it's not on Earth, but on a Jupiter, for instance, the weight is greater than the same mass, right? So g would not be 9.8 meters per second square, but something else. So obviously, on a Jupiter, it requires more because the weight is more so the force should be stronger, the Lorentz force should be stronger and the field should have more intensity. Now, resistance. If resistance is greater than the current is less, right? And obviously the less the current, the less influence the magnetic field has on the wire. I mean, if there is no current at all, there is no Lorentz force. So basically, it should depend on the resistance, the greater the resistance, the more field we need to compensate for the weakness of the current. Now here, in the denominator, we have the voltage. Again, the greater the voltage was given other parameters, the stronger the current and the stronger the current, the less magnetic force we need because the interaction depends on both, magnetic force and the current. They are, you see, proportional to each other. So if this is greater, this can be smaller to compensate. And same thing for the lengths. I mean, the longer the lengths, the more interaction with magnetic field this wire has. Obviously, it's easier. So the greater this, the less this should be. Okay, that's basically the kind of explanation of this formula that really is reasonable. Oh, yes, I forgot about tangent. Now, if the greater the angle is, obviously, the stronger the field should be, right? And the tangent is monotonically increasing function, but in this case from 0 to 90 degrees range. Okay, that's it for the first problem. Now the second problem is related to something the physicists really like to work with. Now right now, or probably during the last 50 years, lots of physicists were involved in elementary particles, which they were using to, using some kind of accelerators, they were bombarding certain other elements with elementary particles, etc. So this is about elementary particles. Now, I will simplify this and then I will explain actually what kind of a relationship it has to elementary particles. Now consider you have a wire and there is no current yet. But there is one specific point charge Q, which is traveling through the wire with the speed B. Now, let's consider that the wire has certain lengths and the field, magnetic field, has certain intensity B. Given. Now my first question is not related to magnetism, it's related to electricity. If one particular charge going like a point charge, like an impulse, if you wish, what is an impulse? Impulse is actually a point charge moving along the wire, right? So if this particular charge is moving along the wire, can we talk about equivalent direct current which basically delivers the same thing, the same result? Now, what is the same result? Certain amount of electricity which is moving during certain amount of time, right? What is the definition of the current? It's amount of electricity in coulombs, let's say, divided by time t per unit of time, right? That's what basically what the definition of the current is. Now, can we talk about equivalent current here? Well, yes, because the time we have, this is L divided by V, distance divided by the speed. Now, the amount of electricity is Q. So when this particular point charge moves along the wire all the way, that's equivalent to I equals 2 Q divided by time, which is L times V. So this is the current which is produced by one single charge moving along a wire, point charge, just by itself. It's not exactly the same as direct current because direct current means a flow of electrons, not just an impulse. However, from the perspective of, you know, practical kind of view to this particular experiment, it looks like we have the current of this kind because, again, we're moving a certain amount of electricity during a certain amount of time. So we can assume, and I definitely recognize that it's not mathematically rigorous assumption. We can assume that instead of one charge, point charge, going through the wire under this condition with the speed and the lengths of the wire, we have a current which is physically equivalent. So we should observe the same kind of results of this experiment as if the current of this kind flowing through the wire. Okay? Again, it's not mathematically rigorous but on the physical level of observing the same results, that's actually true. So what is our problem now? So the first problem was determine the equivalent direct current equivalent to this particular situation. And the second problem is to define to find out what is the Lorentz force exerted on this wire. Well, that's simple. f is equal to i times l times b, right? Which is equal to, l will cancel. So we will have qvb. So if one particular, let's say electron or proton, some electric particle, electrically charged particle, going through the magnetic field, forget about the wire now, l is, it's canceled out. It experiences Lorentz force. It feels this Lorentz force equal to this. It depends on the charge, on the speed and obviously on the field itself. Wire is no longer involved. Why? Well, because this Lorentz force, where is it applicable? If it's a point charge, it's applicable only to this local place where this particular charge is located at one particular moment. So at every moment, the whole force of this magnitude is acting on this charge only, not on the entire wire. So that's what kind of a little deviation, I would say, from the classical view on the Lorentz force. We don't have this wire with a direct current flowing, constantly flowing with the same speed through it. We have one particular particle carrying a charge. And now this force actually acts on this particular particle. So this is the answer to the second problem. This is the Lorentz force, which we wanted to find out. This is the force which the particle actually is acted upon. And this is the second problem. Now the third problem is related to the second problem and I will explain you how. Now let's forget about the wire. We do have a particle and it just flows through the magnetic field. So that's the equivalent of the current going that way and magnetic field going this way. So in this particular case, this particle should deviate from its straight line trajectory because there is a Lorentz force and it's perpendicular to the trajectory and perpendicular to this. So it goes to perpendicular to the board. Now I will change my view. I will look from the top and what do I see in this particular case? The lines of magnetic fields are, that's in this particular case, they go vertically. So all the magnetic field lines, so let's assume that the north is here and the south is behind the board. Now what if I have a particle which is flowing into this magnetic field? Now we are assuming, obviously, that the magnetic field is uniform and it has the intensity b in teslas. So what happens? Well, depending on the direction of the magnetic lines north to south or south to north, let's not think about this right now. Let's consider what will be with the Lorentz force. Lorentz force will be perpendicular to both. To these lines, magnetic lines from north to south and to the current. Well, the particle goes, so this is an electrically charged particle. Q would definitely be, and this is the speed, will definitely be equivalent to a current. And the force will be perpendicular to both, perpendicular to this and perpendicular to this. Which is what? Well, it's a vertical direction. So let's assume it's down. So if it's down, then it should curve. Right? The force goes down. It's perpendicular to this and perpendicular to lines of magnetic field. Now let's just think about it. The force is always perpendicular to the trajectory, which means that at this point, the force is this way. At this point, the force is this way. Then it goes a little further, and the force will be still perpendicular. So it looks like, well, it feels everywhere, magnetic field. It looks like it should actually turn into a circle. Well, let's just think about, is it a circle? Now, and this is a difficult part, because I cannot rigorously prove that this is a circle, because it's kind of involved. There are lots of, you know, calculations and there are some very elegant solutions, but it's a little bit beyond the level of mathematics which I'm presenting in this course. But let me just explain why it should be a circle. Well, first of all, the force, the Lorentz force, is always perpendicular to trajectory. What does it mean? It means that there is no linear acceleration. You see, wherever it goes, if the force is always perpendicular, it does not increase the tangential speed. So there is no linear acceleration. At the same time, the force has always the same value, right? The Lorentz force is equal to what? Q, V, and V, right? From the previous problem, problem number two. So Q is the same. V is a linear speed of this particular charge. And the V is the field, which is uniform. So we always have exactly the same value of force. Direction is always changing perpendicular to trajectory. But the value is the same. So what does it mean? Well, considering we have a mass, which is always the same, so we have a force, we have a mass, so there is second Newton's law, right? Which gives me an acceleration perpendicularly to the trajectory. Do you remember the rotation? The kinematics of rotation is that there is always centripetal acceleration A, which is equal to what? r omega square or r or V square divided by r. Omega is angular speed and V is linear speed, the same as this one. So we will use this particular thing. So A is a centripetal acceleration always towards the center and always perpendicular to the trajectory, right? Exactly the same as in our case. The force is perpendicular. The Lorentz force is perpendicular to trajectory, right? And its constant value, constant absolute value. So it really looks like this circular movement satisfies completely our conditions. Now, from the physical standpoint, it's usually sufficient to say, okay, that's why the trajectory is a circle. As a mathematician, I would say that's not exactly a rigorous proof because maybe there is some kind of other curve in the plane, which also has the same properties that the linear speed would be constant and acceleration perpendicular to the linear speed would also be constant, but not a circle. Well, and this is exactly what the difficult part to prove that there is no such curve is not easy. So just take my word and there are actually a couple of websites, which maybe I will put into the notes for this lecture, that basically goes to a rigorous proof that the circle is the only thing which is possible. So we assume that circle is the trajectory. So circle trajectory definitely satisfies all our parameters. Now, what is my question? My question is what is the radius of this given whatever else parameters we need? Well, the radius we can define from here, right? r is equal to v squared divided by a. We know v, okay? a, well, a is f divided by m. That's the second Newton's law. And f we also have. So what do we have? We have r is equal to v squared divided by f, which is q, v, b, and m goes to the top. So b, m goes to the top, which is equal to mv divided by qb. mv is by the way momentum of motion. So the greater the momentum of motion, the greater the radius will be. So either the mass is very big or the speed is very big. And to turn it needs obviously greater radius. It will go more greater radius if it's stronger momentum. Now, if charge is greater or intensity of the field is greater, the force will be greater, which actually rounds up the trajectory. So that's basically the kind of reasonable formula. And that's how they calculate what exactly kind of accelerator we need if we would like to, let's say, accelerate one particular proton, for instance. So we know the q, the charge of the proton. We know the mass of the proton. And depending on what kind of a speed we need and what kind of an intensity of the magnetic field we can do, this will give us the radius. And the radius can be very big. I mean, there are some accelerators which are miles in the diameter, right? So that gives you the radius, how to build an accelerator, a cyclotron, right? Okay, that's it for today. I do recommend you to read my notes for this lecture. Good luck. Thank you.